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How do you rearrange a literal equation or formula to solve for a specified variable?

Rearrange formulas and literal equations to solve for a specified variable, treating the other letters as constants and using inverse operations (A.EI.1).

A Virginia SOL Algebra I answer on rearranging literal equations and formulas: isolating a chosen variable, treating other letters as constants, clearing fractions, and factoring out the target variable when it appears twice.

Generated by Claude Opus 4.89 min answer

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Jump to a section
  1. What this topic is asking
  2. The method: same moves, letters instead of numbers
  3. Clearing fractions and coefficients
  4. When the variable appears twice
  5. Why rearranging a formula is the same skill as solving
  6. How the SOL examines this topic
  7. Try this

What this topic is asking

This part of A.EI.1 asks you to rearrange a formula or literal equation to solve for a specified variable. A literal equation is one with several letters (like A=12bhA = \frac{1}{2}bh or Ax+By=CAx + By = C), and you isolate the one the item names, treating the others as if they were numbers. On the Virginia Algebra I SOL these are Equations and Inequalities items, usually fill-in-the-blank (type the rearranged formula) or multiple choice.

The method: same moves, letters instead of numbers

Solving a literal equation uses the same balance method as a one-variable equation. The only difference is that the answer is an expression in the other variables, not a number. The key mental move is to decide which letter is the unknown and treat all the others as constants.

For P=2l+2wP = 2l + 2w, solving for ww:

P=2l+2w  ⟹  P−2l=2w  ⟹  w=P−2l2.P = 2l + 2w \implies P - 2l = 2w \implies w = \frac{P - 2l}{2}.

Subtract the 2l2l term (it does not contain ww), then divide the whole side by 22.

Clearing fractions and coefficients

When the target variable is multiplied by a fraction or divided by something, clear it first.

When the variable appears twice

If the target variable shows up in more than one term, you cannot isolate it by a single division. Collect those terms on one side and factor the variable out, then divide by the bracket.

For example, solve ax+b=cx+dax + b = cx + d for xx: move the xx terms together, ax−cx=d−bax - cx = d - b, factor, x(a−c)=d−bx(a - c) = d - b, then divide, x=d−ba−cx = \frac{d - b}{a - c}. The factoring step is what makes a single division possible.

Why rearranging a formula is the same skill as solving

A literal equation feels different because of the letters, but algebraically nothing new is happening. When you solve 3x+7=223x + 7 = 22, you subtract 77 and divide by 33; when you solve bh=2Abh = 2A for hh, you divide by bb. In both cases you apply inverse operations to both sides to peel away everything attached to the target. The letters AA and bb behave like the numbers 2222 and 33 would: they are simply known quantities you have not been given values for. This is why the topic matters for the rest of Algebra I: rearranging Ax+By=CAx + By = C into y=mx+by = mx + b form is exactly how you graph a line given in standard form, and solving a formula for a variable is how you prepare it to be evaluated repeatedly. Recognizing rearrangement as ordinary equation solving (with symbols standing in for numbers) removes the mystery.

How the SOL examines this topic

  • Fill-in-the-blank. Solve a given formula for a named variable and type the expression.
  • Multiple choice. Pick the correct rearrangement, with distractors that divide only part of a side or reverse a subtraction.
  • Drag-and-drop. Order the steps of an isolation, or assemble the rearranged formula.

A clarifying idea: the answer to "solve for yy" is an expression, so it will still contain the other letters. If your answer has no other variables left, you probably treated a letter as something to eliminate rather than as a constant.

Try this

Q1. Solve d=rtd = rt for tt. [1 point]

  • Cue. Divide by rr: t=drt = \frac{d}{r}.

Q2. Solve 2x+3y=122x + 3y = 12 for yy. [2 points]

  • Cue. 3y=12−2x⇒y=12−2x33y = 12 - 2x \Rightarrow y = \frac{12 - 2x}{3}.

Exam-style practice questions

Practice questions written in the style of VDOE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SOL (style)2 marksFill in the blank. The area of a triangle is A=12bhA = \frac{1}{2}bh. Solve for hh in terms of AA and bb.
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The result is h=2Abh = \dfrac{2A}{b}.

Treat AA and bb as constants and isolate hh. Multiply both sides by 22 to clear the fraction: 2A=bh2A = bh. Divide both sides by bb: h=2Abh = \frac{2A}{b}. A common error is dividing by 22 instead of multiplying, which would put the 22 in the denominator. Undo the times-one-half by multiplying by 22.

SOL (style)1 marksMultiple choice. The equation of a line is Ax+By=CAx + By = C. Which expression gives yy? (A) y=C−AxBy = \dfrac{C - Ax}{B} (B) y=C−Ax−By = C - Ax - B (C) y=CB−Axy = \dfrac{C}{B} - Ax (D) y=Ax−CBy = \dfrac{Ax - C}{B}
Show worked answer →

The correct answer is (A).

Isolate the yy term: subtract AxAx from both sides to get By=C−AxBy = C - Ax. Divide every term on the right by BB: y=C−AxBy = \frac{C - Ax}{B}. Option (C) divides only the CC by BB, not the whole right side; option (D) has the subtraction reversed. The whole numerator must be divided by BB.

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